Suppose that a van with capacity for 7 passengers departs from a commuter station

Question: Suppose that a van with capacity for 7 passengers departs from a commuter station. Observation has shown that the bus never departs empty. Let X denote the number of passengers that are female and Y the number of passengers that are male. Assume that all possible (x, y) pairs are equally likely. (a) Find the probability that the number of female passengers is more than 5. (a) Find the probability that the total number of passengers is less than 5.

Solution:

pairs of passengers:-
{(0,7),(0,6),(0,5),(0,4),(0,3),(0,2),(0,1)
(1,6),(1,5),(1,4),(1,3),(1,2),(1,1),(1,0)
(2,5),(2,4),(2,3),(2,2),(2,1),(2,0)
(3,4),(3,3),(3,2),(3,1),(3,0)
(4,3),(4,2),(4,1),(4,0),
(5,2),(5,1),(5,0)
(6,1),(6,0)
(7,0)}
p(x,y) = 1/35 where x = no of female and y = no of males

a. Probability of female passengers is more than 5
possible outcomes = (6,0),(6,1),(7,0)
P(A) = 3/35 = 0.0857

b. Passenger is less than 5
possible outcomes = {(4,0),(3,0),(3,1),(2,2),(2,1),(2,0),
(1,3),(1,2),(1,1),(1,0),(0,3),(0,2),(0,1),(0,4)}
= 14
P(B) = 14/35 = 0.4

Which statement below about asexual reproduction is FALSE?

Question: Which statement below about asexual reproduction is FALSE?
A. With asexual reproduction, offspring are genetically equivalent to the parent.
B. Asexual reproduction requires no partner.
C. Asexual reproduction requires meiosis.
D. Some organisms can reproduce both sexually and asexually.

Solution: Asexual reproduction requires meiosis.
only sexual reproduction required meiosis, as gamete fusion occurs, to maintain the stable chromosome number in the offspring due to this phenomenon, meiosis is required in sexual reproduction.

Question: If a diploid sperm fertilized a diploid egg, what would the result be?
A. Diploid (2n)
B. Quadruploid (4n)
C. A mixture of diploid (2n) and quadruploid (4n)
D. Triploid (3n)

Solution: The answer is Quaadruploid (4n)
2n+2n = 4n.

Question: Which of the following represents things that are equivalent?
A. Two alleles for the same gene in a homologous chromosome pair
B. The sequences of DNA in the two sister chromatids of a chromosome before meiosis
C. The sequences of DNA in the two sister chromatids of a chromosome after meiosis
D. The number of homologous chromosomes per cell before and after meiosis

Solution: Two alleles for the same gene in a homologous chromosome pair.

Question: What problem would most likely occur if a haploid cell attempted to perform meiosis?
A. The cell could not replicate its DNA prior to meiosis.
B. The cell could not pair homologous chromosomes during meiosis I.
C. The cell would produce diploid daughter cells.

Solution: The cell could not pair homologous chromosomes during meiosis I.

Question: Which statement below is TRUE?
A. Crossing over shuffles genes between non-homologous chromosomes.
B. During melosis, cells arrange their alleles so that beneficial alleles are passed on together.
C. Without crossing over, offspring would be genetically identical to parents.
D. New combinations of alleles arise from the random events of crossing over, independent assortment, and mutation.

Solution: New combinations of alleles arise form the random events of crossing over, independent assortment, and mutation.

Question: What statement below is TRUE of the spindle during mitosis and meiosis?
A. The spindle always separates sister chromatids during anaphase.
B. The spindle always separates homologous chromosomes during anaphase.
C. Chromosomes are always attached to both spindle poles during metaphase.
D. The spindle always attaches to chromosomes at the kinetochore.

Solution: The spindle always attached to chromosomes at the kinetochore.

Question: If a person is heterozygous for the △32 allele of the CCR5 gene, how many of the four daughter cells produced by meiosis will have the △32 allele?
A. 1
B. 2
C. 4
D. Varies

Solution: 2.

Question: Which of the following processes has the effect of changing ploidy?
A. DNA replication
B. Meiosis I
C. Meiosis II
D. Mitosis

Solution: Meiosis I
During Anpahse I of meiosis I, as the chromosome segregation occurs, the ploidy will be reduced to half.

Question: The image below shows a homologous chromosome pair with two genes, A and B. Each of the two genes has two alleles (A1 and A2, B1 and B2). For this homologous pair of chromosomes, after completion of meiosis with NO CROSSING OVER, which of the following combinations of alleles might end up in one of the gametes?


A. A1 and B1
B. A2 and B1
C. A2 and B2
D. All of the above are possible.

Solution: A2 and B1

Question: For the same homologous pair of chromosomes shown in the previous question, which of the following combinations of alleles might end up in one of the gametes after completion of meiosis WITH CROSSING OVER?
A. A1 and B2
B. A2 and B1
C. A2 and B2
D. All of the above are possible

Solution: All of the above are possible.

Describe the physical changes that occur during frying hen eggs temperature 180°C

Question: 1. Describe the physical changes that occur during frying hen eggs temperature 180°C? at

2. Discuss how nitrites can be used to preserve cured meat products.

Solution:

Ans 1 Deep frying is an ancient process commonly used in food industry. It is a lost cost and highly demanding process which involves oil-food interaction at high temperature. When food is cook at high temperature dehydration occurs and food gets completely cooked because of which physical and chemical changes take place. There are vareity of examples such as starch gelatinization, protein denaturation, caramelization, maillard reaction ( giving food an aromatic and mouth watering aroma, which occurs when amino acids and sugar breaks down). Similarly deep frying of hen egg involves protein denaturation. The protein is a three dimensional arrangement of atoms in amino acid chain molecules and fold into spatial conformations via non colent interactions such as hydrogen bonding, ionic interactions and vanderwall forces. When the whole egg i.e yolk and white( albumen-major source of protein) is heated it turn into solid. This solidification of protein is known as coagulation. The heat causes the protein structure to lose water molecules and hence shrink them. Different proteins denature at diiferent temperature, similarly egg white starts denaturing at 61 degree celcius. As the protein changes its shape, it changes its taste and texture also.

Ans 2 Salts ( sodium chloride) are used as preservatives for meats from a long time.It was accidently found that nitrite salts increases the shelf life of meats by preventing rancidity( spoilage of food) and also the growth of bacteria. So now a days nitrite salts  such as sodium nitrites are being used as preservatives and purified nitrites are commercially manufactured for this purpose. Pure nitriles if consumed at levels of 3-5 gm can be dangerous and may result in death also as it binds with oxygen in body stronger than heamoglobin thus not allowing oxygen to reach the organs. So it is added to meat according to level(0.01-0.02% depending on type of meats) set by USDA. It helps in improving meat quality , when added at allowed levels set by USDA it inhibits the growth of almost all bacteria such as Clostridium botulinum, Clostridium perfringens, Listeria monocytogens.that caauses food spoilage and sickness.

Please implement booth’s algorithm in logisim to solve signed multiplication.

Question:

Please implement booth’s algorithm in logisim to solve signed multiplication. The circuit should represent the one seen below.  

The following inputs and outputs are required:

• Multiplicand: a 16-bit two’s complement input

• Multiplier: a 16-bit two’s complement input

• Mul (1-bit input): This input will be one if the instruction is the multiplication instruction

• Clock (1-bit input)

• Product: 32-bit two’s complement output

• M Ready (1-bit output): This output will be 1 if the product is ready

The inside circuit should use a register for the multiplier, the multiplicand, and the product.  A ROM circuit should control the clock signal and what happens on each clock edge. The ROM circuit triggers the registers. The ROM circuit can be seen below:

Solution:

Booth’s algorithm is a multiplication algorithm that multiplies two signed binary numbers in 2’s compliment notation.-

Determine P(A1 I B1) and Determine P(A2 I B1)

Question:

       A1     A2

B1 .4      .3

B2 .2      .1

Determine P(A1 I B1)

Determine P(A2 I B1)

Did your answers to parts (a) and (b) sum to 1? Is this a coincidence? Explain

Solution: a. P(A1 I B1) = P( A1 n B1)/ P(B1)

P( A1 n B1) = 0.4

P(B1) = 0.4+0.3 = .7

P(A1 I B1) = 0.4/0.7 = 0.5714

b. P(A2 I B1) = P( A1 n B1)/ P(B1)

P( A2 n B1) = 0.3

P(B1) = 0.4+0.3 = .7

P(A1 I B1) = 0.3/0.7 = 0.4285

Tangent Lines Find the equation of the tangent line of the curve tan(2x+5y) = x at (0,0).

Question: Calculus: Tangent Lines Find the equation of the tangent line of the curve tan(2x+5y) = x at (0,0). Pick ONE option

y = 5x

y=x

y = x/5

y=-x/5

Solution:

We have to find the equation of the tangent line of the curve

tan(2x+5y)=x

at the point (0,0).

Now, to find this, we have to find the value of

which is the slope of the tangent line, at a given point.

Here,

Differentiating both sides with respect to x, we get

Putting the value of x=0 and y=0, we can say that

So, the slope of the tangent will be -(1/5).

Now, at the point (0,0), the equation of the straight line is

So, the correct answer is option (D) .

The double riveted butt joint shown in the Figure connect two plates, which are each 2.5 mm thick

Question: The double riveted butt joint shown in the Figure connect two plates, which are each 2.5 mm thick, the rivet has a diameter of 3mm. If the failure strength of the rivets in shear is 370 N/mm2 and the ultimate tensile strength of the plate is 465 N/mm2 determine the necessary rivet pitch if the joint is to be designed so that failure due to shear in the rivets and failure due to tension in the plate occur simultaneously. Calculate the joint efficiency.

Solution:

Given that:
Thickness = 2.5 mm
Diameter = 3 mm
Tensile strength = 465 N/mm^2
Shear strength = 370 N/mm^2
Shearing of rivets can be calculated as below:
Ps = nπ/4 * d^2 * τ * 2 (double shear)
where
τ = shear strength
n = no of rivets pen pitch length = 2
d = 3
Now substitute in above

Ps = 2 π/4 * 3^2 * 370 * 2
Ps= 10461.5 N

Tearing resistance of plate:
Pt = (P-d)tθt
d = 3 mm
t = 2.5 mm
θt = 465 N/mm^2

Pt = (P – 3) * 2.5 * 465
As given question shearing resistance of rivet = tearing resistance of plate
that is
Ps = Pt
(P-3)2.5465 = 10462
(P-3) * 1162.5 = 10462
P – 3 = 10462/1162.5
P – 3 = 8.999
P = 8.99 + 3
P = 11.99
P = 12
Now the strength of the unriveted on solid plate pen pitch length
P = Ptθt
t = 2.5 mm
P = 12
θt = 465
Now substitute those value’s then
P = 122.5465
P = 13950N
where P= pitch of revert
Now to determine the efficiency of the joint we know that the
efficiency of joint = least of Ps,Pt/P
Now substitute those values
Ps = 10462N
P = 13950
= 10462/13950
= 0.749
Efficiency of joint = 74.9%

The Company has purchased a 1-year software license for the use of the Sales team from Widgets

Question: The Company has purchased a 1-year software license for the use of the Sales team from Widgets, Inc. The total cost of the license is $5,000, and it is valid from July 25, 2020 through July 24, 2021. The invoice is dated August 1, 2021, and due within 30 days. Please write the entry or entries you would use to record this transaction. Include debit(s), credit(s), and a memo for each entry.

Solution:

Journal Entries
DateAccount Titles and ExplanationDebitCredit
Jul. 25, 2020Software License$5,000
   Accounts Payable$5,000
(To record the purchase of software license on account)
Jul. 24, 2021Accumulated Amortization – Software License$5,000
   Software License$5,000
(To record the expiry of software license)
Aug. 31, 2021Accounts Payable$5,000
   Cash$5,000
(To record the payment for purchase of software license)

Group A Question 1 The two port circuit presented in Fig. 1 can be described in terms of b-parameters as:

Question: Group A Question 1 The two port circuit presented in Fig. 1 can be described in terms of b-parameters as:

Solution:

It can be proved by advanced circuit theory that voltages and currents in Fig. can be related

by the following sets of equations :

v1 = h11 i1 + h12 v2 …(i)

i2 = h21 i1 + h22 v2 …(ii)

In these equations, the hs are fixed constants for a given circuit and are called h parameters.

Once these parameters are known, we can use equations (i) and (ii) to find the voltages and currents

in the circuit. If we look at eq.(i), it is clear that **h11 has the dimension of ohm and h12 is dimension-

less. Similarly, from eq. (ii), h21 is dimensionless and h22 has the dimension of mho. The following

points may be noted about h parameters :

(i) Every linear circuit has four h parameters ; one having dimension of ohm, one having di-mension of mho and two dimensionless.

(ii) The h parameters of a given circuit are constant. If we change the circuit, h parameters would also change.

The major reason for the use of h parameters is the relative ease with which they can be measured. The

h parameters of a circuit shown in Fig. 24.1 can be found out as under :

(i) If we short-circuit the output terminals (See Fig. 24.2), we can say that output voltage v2 = 0.

Putting v2 = 0 in equations (i) and (ii), we get,

v1 = h11 i1 + h12 × 0

i2 = h21 i1 + h22 × 0

∴ h11 = v1/i1 for v2 = 0 i.e. output shorted

and

h21 = i2/ i1 , for v2 = 0 i.e. output shorted

Let us now turn to the physical meaning of h11 and h21 . Since h11 is a ratio of voltage and current

(i.e. v1/i1), it is an impedance and is called * “input impedance with output shorted ”. Similarly, h21 is the ratio of output and input current (i.e., i 2/i 1), it will be dimensionless and is called“current gain with output shorted”.

The other two h parameters (viz h12 and h22) can be found by making i

1 = 0. This can be d doneby the arrangement shown in Fig.. Here, we drive the output terminals with voltage v2,

keeping the input terminals open. With this set up, i

1 = 0 and the equations become :

v1 = h11 × 0 + h12 v2

i 2 = h21 × 0 + h22 v2

∴ h12 = v1/v2 for i1 = 0 i.e. input open

and

h22 = i2 /v2 , for i1 = 0 i.e. input open

Since h12 is a ratio of input and output voltages (i.e. v1/v2), it is dimensionless and is called “voltage

feedback ratio with input terminals open”. Similarly, h22 is a ratio of output current and output voltage

(i.e. i 2/v2), it will be admittance and is called “output admittance with input terminals open”.

Find the number of pumps required to take water from a deep well under a total head of 120 m

Question: Find the number of pumps required to take water from a deep well under a total head of 120 m. All the pumps are identical and are running at 800 rpm. The specific speed of each pump is given as 25 while the rated capacity of each pump is 0.16 m3/s

Solution: Given: total head h = 120 m
Speed, N = 800 rpm
Specific speed at each pump, Ns = 25
Rated capacity of each Pump, Q = 0.16 m^3/s
Let us assume, head developed by each pump is Hm
Specific speed, Ns = N (Q)^(1/2)/ (Hm)^(3/4)
25 = 800(0.16)^(1/2)/(Hm)^(3/4)
(Hm)^(3/4) = 800 (0.16)^(1/2)/25
(Hm)^(3/4) = 12.8
Hm = (12.8)^(4/3)
head developed by each pump
Hm = 29.94 m

Number of pumps required
= total head/head developed by each pump
= 120/29.94 = 4.008 = 4
4 numbers of pumps are required to lift water from deep well under a total of 120m